Simplify and expand the following expression: $ \dfrac{3}{2r - 12}- \dfrac{1}{5r + 50}- \dfrac{1}{r^2 + 4r - 60} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{3}{2r - 12} = \dfrac{3}{2(r - 6)}$ We can factor a $5$ out of denominator in the second term: $ \dfrac{1}{5r + 50} = \dfrac{1}{5(r + 10)}$ We can factor the quadratic in the third term: $ \dfrac{1}{r^2 + 4r - 60} = \dfrac{1}{(r - 6)(r + 10)}$ Now we have: $ \dfrac{3}{2(r - 6)}- \dfrac{1}{5(r + 10)}- \dfrac{1}{(r - 6)(r + 10)} $ The least common multiple of the denominators is: $ 10(r - 6)(r + 10)$ In order to get the first term over $10(r - 6)(r + 10)$ , multiply by $\dfrac{5(r + 10)}{5(r + 10)}$ $ \dfrac{3}{2(r - 6)} \times \dfrac{5(r + 10)}{5(r + 10)} = \dfrac{15(r + 10)}{10(r - 6)(r + 10)} $ In order to get the second term over $10(r - 6)(r + 10)$ , multiply by $\dfrac{2(r - 6)}{2(r - 6)}$ $ \dfrac{1}{5(r + 10)} \times \dfrac{2(r - 6)}{2(r - 6)} = \dfrac{2(r - 6)}{10(r - 6)(r + 10)} $ In order to get the third term over $10(r - 6)(r + 10)$ , multiply by $\dfrac{10}{10}$ $ \dfrac{1}{(r - 6)(r + 10)} \times \dfrac{10}{10} = \dfrac{10}{10(r - 6)(r + 10)} $ Now we have: $ \dfrac{15(r + 10)}{10(r - 6)(r + 10)} - \dfrac{2(r - 6)}{10(r - 6)(r + 10)} - \dfrac{10}{10(r - 6)(r + 10)} $ $ = \dfrac{ 15(r + 10) - 2(r - 6) - 10} {10(r - 6)(r + 10)} $ Expand: $ = \dfrac{15r + 150 - 2r + 12 - 10}{10r^2 + 40r - 600} $ $ = \dfrac{13r + 152}{10r^2 + 40r - 600}$